Collision

1. Collision Reporters: CabagAndaluz Diaz Janier

2. CRASHES BILLIARDS

3. The act or process of colliding; a crash or conflict. A brief dynamic event consisting of the close approach of two or more particles, such as atoms, resulting in an abrupt change of momentum or exchange of energy.

4. Momentum • Involves motion -to have momentum, an object must be moving at a certain velocity • Involves mass -the more massive, the more difficult to change your state immediately -is the measure of one’s motion, equivalent to the product of one’s mass and velocity.

5. Express in: p= m v Where p is the momentum ain kg-m/s

6. Sample conditions: • If you have a VW Bug going 50 km/h and a truck going at the same velocity, which has more momentum? • What about two trucks, one moving at 25 km/h and the other moving 50 km/h. Which has more momentum?

7. Example: Calculate the momentum of 110 – kg missile traveling at 100 m/s eastward. Given: Mass (m) – 110 kg of missile Velocity (v) – 100 m/s Momentum - ?

8. Solution: Direct substitution p = mv = 110 kg (100 m/s) = 11000 kg – m/s, eastward

10. Impulse • Force (F) applied by one object to another object within a given time interval ∆t Express in: F∆t= ∆p Where F∆t is the impulse in newton-second (N-s)

11. Derivation: From Newton’s Law: F=m a Since acceleration is: F = m

12. Transporting ∆t to the other side of equation: F ∆t = m ∆v F ∆t= ∆p

13. Conservation of Momentum • All collisions have to conserve momentum.

15. Elastic Collision in One Dimension

16. =

17. since ( /

18. Case 1: Both masses are equal (

19. Stops

20. Case 2: Mass of moving object greater than the mass of the stationary object () for

21. Example (case 1) Before collision After collision

22. Example (case 2) 2 kg 1 kg 1 m/s = (1/3) m/s 1 m/s = (4/3) m/s

23. Before collision 2kg 1 m/s 1 kg After collision (1/3) m/s 2/3 m/s

24. CASE 3

25. Mass of moving object less than the mass of the stationary object (m1 < m2)(m1 - m2) v’ = (m1 + m2) v1 m1 Moves. But, in the opposite direction.2m1v’=(m1 + m2) v1m2 Moves. V2 has the same direction as V1

26. Example : Again, m2 is stationary on the track while the other m1 rolls at 1 m/s towards the right. This time, the mass of m2 is 2 kg while the mass of m1 is 1 kg. What happens when the balls collide elastically?

27. Since the masses are not equal, we solve again for v1’ and v2’ So(m1 - m2) 1 kg – 2 kg v’ = (m1 + m2) v1 = 2 kg + 1kg = 1 m/s = -(1/3) m/s NEGATIVE OPPOSITE DIRECTION2m1 2 kgv2’=(m1 + m2) v1 = 3 kg = 1 m/s = (2/3) m/s

28. Before Collisionm1 1m/s m2 1kg 2kgAfter Collisionm1m2 (1/3) m/s 2/3 m/s

29. Total inelastic collision in one dimension

30. If in an elastic collision the masses maintain their mass state after collision, a total inelastic collision produces a combined mass after collision

31. m1 m2 Before Collision 1 m/sAfter Collision VfFormula:m1v1 + m2v2 = (m1 + m2) vf m1 + m2

32. If m2 is stationary, m2v2 = 0 FORMULA: m1Vf = v1 (m1 + m2)

33. Example: Colliding with a much heavier object may produce an inelastic collision. A head-on collision between a car and a train may drag the car along the path of the train. Assume that 1000-kh car stalled on the railroad tracks is smashed by a 50,000-kg train travelling at 8.33 m/s (about 30 km/hr). What is the final velocity after a total inelastic collision?

34. m1Vf = v1 (m1 + m2) = 8.33 m/s 50,000 kg 51,000 kg = 8.17 m/s

35. Collisions in two dimensions

36. When we move to collisions in two dimensions, we can separately balance momentum in the x-direction and in the y-direction Before After p1’ p2’ a1a2 ß1 ß2 p1 p2

37. Using the conservation of energy: Along the x-axis: ∑pxbefore collision = ∑pxafter collision (Eqn. 7.16)p1cos a1 + p2cos a2 = p1’ cos ß1 + p2’ cos ß2 Along the y-axis: ∑pybefore collision = ∑pyafter collision (Eqn. 7.17)p1 sin a1 + p2 sin a2 = p1’ sin ß1 + p2’ sin ß2

38. Example A pool ball weighing 2 kg is traveling at 30◦ at 0.8 m/s hits another ball moving at 0.5 m/s at 180◦. If the second ball leaves the collision at 0◦ and the first moves away at 150◦, find the final velocity vectors of the balls. Before After v’ 1 0.5 m/s ß1 0.8 m/s v’ 2 30◦ ß2 = 0

39. Answer Given mass of the pool balls m1, m2 2kg Before Collision velocity of the first pool ball v1 0.8m/s angle of the first pool ball a1 30◦ Velocity of second pool ball v2 0.5 m/s Angle of second pool ball a2 180◦ After collision Angle of first pool ball ß1 150◦ Angle of second pool ball ß2 0◦ Find: Final velocities of the balls ( v1’ and v2’)

40. Solution Along the x-axis: ∑px before collision = ∑px after collision p1cos a1 + p2cos a2 = p1’ cos ß1 + p2’ cos ß2 m(v1cos a1 + v2cos a2) + m(v1’ cos ß1 = v2’ cos ß2) v1cos a1 + v2cos a2 = v1’ cos ß1 + v2cos ß2 0.8 m/s (cos 30◦) + 0.5 m/s (cos 180◦) = v1’ cos 150◦ + v2’ cos 0◦ 0.8m/s (0.866) + 0.5 m/s (-1) = v1’ (-0.866) + v2’ (1) 0.70 m/s – 0.5m/s = -0.866 v1’ + v2’ Expressing v2’ in terms of v1’ v2’ = 0.866 v1’ + 0.20m/s

41. Along the y-axis: ∑py before collision = ∑pyafter collision p1 sin a1 + p2 sin a2 = p1’ sin ß1 + p2’ sin ß2 m( v1 sin a1 + v2 sin a2) = m(v1’ sin ß1 + v2’ sin ß2) v1 sin a1 v2 sin a2 = v1’ sin ß1 + v2’ sin ß2 0.8 m/s (sin 30◦) + 0.5m/s (sin 180◦) = v1’ sin 150◦ + v2 sin 0◦ 0.8m/s (0.5) + 0.5m/s (0) = v1’ (0.5) + v2’ (0) 0.4m/s = 0.5 v1’ 0.4m/s / 0.5 = v1’ v1’ = 0.m/s < 150◦ Solving for v2’ v2’ = 0.866 v1’ + 0.20 m/s = 0.866 (0.8m/s) + 0.20 m/s = 0.9 m/s <0◦